Answer
(a) Speed of the windis , $v_w=44.72 \space km/h$, $63.34^0$ north of west.
(b) Direction of motion of airplane w.r.t. air is $10.48^0$ north of west.
Work Step by Step
(a) Velocity of airplane (p) with respect to air (a), $v_{pa}=-(220 \space km/h) \hat{i}$.
The displacement of the airplane w.r.t. ground, $s_{p}=-(120 \space km) \hat{i}-(20 \space km) \hat{j}$.
So, velocity of airplane (p) w.r.t. ground will be, $v_{p}=\frac{s_p}{t}=\frac{-(120 \space km) \hat{i}-(20 \space km) \hat{j}}{0.5\space h}$
$\Rightarrow v_p=-(240 \space km/h) \hat{i}-(40 \space km/h) \hat{j}$
Now, $v_{pa}=v_{p}-v_{a}\Rightarrow v_{a}=v_{pa}-v_{p}=-(20 \space km/h) \hat{i}+(40 \space km/h) \hat{j}$
So, speed of the windis , $v_w=\sqrt{(-20)^2+40^2}=20\sqrt5 \space km/h=44.72 \space km/h$
And its direction is, $\theta=tan^{-1}|{\frac {v_y}{v_x}|}=tan^{-1}2=63.34^0$ north of west.
(b) Let speed of the airplane w.r.t. air $220 \space km/h$ at an angle $\theta$ towards north of west.
Then $v_{pa}=-220cos\theta \hat{i}+220sin\theta \hat{j}$
Now, velocity of airplane w.r.t. ground, $v_{p}=v_{pa}+v_{a}=-220cos\theta \hat{i}+(220sin\theta-40) \hat{j}$
For airplane to travel westward, y-component of $v_p$ must be zero,
So, $220sin\theta-40=0\Rightarrow sin\theta=\frac{40}{220}\Rightarrow theta=sin^{-1}\frac{2}{11}=10.48^0$ north of west.
